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Homework Help: Differential equation problem

  1. Sep 1, 2008 #1
    1. The problem statement, all variables and given/known data

    Dear Friends,

    Given the differential equation

    [tex]\frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)[/tex]

    with the condition that x(0) = 0

    Then find the largest possible solution (this is how its stated) if either

    [tex]\alpha_{1} = 1/2[/tex]

    [tex]\alpha_{2} = 1[/tex]

    3. The attempt at a solution

    Don't I treat the above as a seperable differential equation?

    Such that

    [tex]{(\frac{1}{1+x^2})}^{-1})} \frac{dx}{dt} = {({\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)})}^{-1}[/tex]

    [tex]1+x^2 dx = \frac{2}{\alpha \cdot \pi \cdot cos(t)} dt[/tex]

    [tex]\int x^2 dx = \int \frac{2}{\alpha \cdot \pi \cdot cos(t)} -1 \ dt + C[/tex]

    [tex] \frac{x^3}{3} = \frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi}[/tex]

    [tex] x^3 = 3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})[/tex]

    [tex] x(t) = (3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})^{\frac{1}{3}}[/tex]

    if I then insert x(0) = 0.

    I get C = 0. Then this means that max solution is 0??

    I just want to be sure. Can anybody please look at my result at then inform me. Have done this correctly?

    Thank You in Advance

  2. jcsd
  3. Sep 1, 2008 #2


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    Homework Helper

    \frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)

    [tex]\Rightarrow \frac{1}{1+x^2}dx=\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t) dt[/tex]

    Now just integrate both sides.
  4. Sep 1, 2008 #3
    I then get

    [tex]tan(x)^{-1} = \frac{\alpha \cdot 2 \pi}{4} \cdot sin(t) [/tex]

    I then take arctan on both sides of equation to arrive at a solution? with regards to the two alpha's?
  5. Sep 1, 2008 #4
    I meant to I take tan on each side of the equation, and thusly obtain x(t) = ??

    and then by choosing either alpha1 or alpha2 see which of these gives largest possible solution???
  6. Sep 2, 2008 #5


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    Science Advisor

    You mean tan-1(x) or arctan(x). There is no need to solve for x(t). Just differentiate with respect to [itex]\alpha[/itex] to find a maximum.
  7. Sep 2, 2008 #6
    Dear Mister Hallsoft,

    Just to be clear I find the derivative with respect to alpha of the original expression

    [tex]\frac{1}{(x^2+1)} dx = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t) dt[/tex]

    [tex]\frac{-2x}{x^4 + 2x^2 +1} = \frac{\pi \cdot cos(t)}{4}[/tex]

    and then insert the two alpha values to obtain the maximum??

    Best Regards

    Last edited: Sep 2, 2008
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